How to prove that $\mathrm{int}(f(U))=\varnothing$ Let $f:U\subset\mathbb{R}^m\longrightarrow\mathbb{R}^n$ be a continuous injective function where $m<n$ and $U$ an open set.
How to prove that $$\mathrm{int}(f(U))=\varnothing$$
Any hints would be appreciated.
 A: As $U$ is open in $\mathbb R^m$ we can writte $U=\bigcup_{k=0}^\infty Q_k$ where $\{Q_k\}_{k=0}^\infty$ is an almost disjoint family of cubes; by almost disjoint I mean $int(Q_i)\cap int(Q_j)=\emptyset$ whenever $i\neq j$. 
As each $Q_i$ is compact and $f$ is injective, we obtain that each $f\upharpoonright Q_i$ is an embedding of $Q_i$ into $\mathbb R^n$, hence $f[Q_i]$ and  $Q_i$ are homeomorphic.
Let us see $int(f[Q_i])=\emptyset$. Suppose not, then there would be some ball $B$ of $R^n$ such that $B\subseteq f[Q_i]$, but as $f[Q_i]$ is homeomorphic to $Q_i$ we get that $ind(f[Q_i])=m$ ; where $ind(X)$ denotes the inductive dimension of a topological space $X$, however $B\subseteq f[Q_i](\clubsuit)$, thus $ind(B)\leq ind(f[Q_i])$, but $ind(B)=n$, contradiction.
We have $f[U]=\bigcup_{k=0}^\infty f[Q_i]$, hence by Baire's category theorem we obtain that $int(f[U])=\emptyset$; as each $f[Q_i]$ is a closed nowhere dense subset of $\mathbb R^n$.

$(\clubsuit)$It is not true in general that if $M$ is a subspace of  space $X$ we have $ind(M)\leq ind(X)$, however, this does hold if $X$ is regular; check Theorem 7.1.1. of Engelking's General Topology.
